Statistics 1L03 - Midterm # Review Atinder Bharaj Made with L A TEX October, 01 Introduction As many of you will soon find out, I will not be holding the next midterm review. To make it a bit easier on students and myself, I will be constructing a set of questions from the textbook that will focus on preparing you for the next midterm. These are adequate questions, so hopefully you learn something from them and they help. 5. These questions below will focus on word problems mainly. However, unlike section 5.5, we can have multiple cases in these problems. To tell if there are cases in a question, look for words like at least... or at most... Question 1 (#, pg 5) Selecting balls from an Urn: An urn contains 15 numbered balls, of which are red and the rest are white. A sample of balls is to be selected. a) How many different samples are possible? Since order won t matter in this question (you can t tell the same coloured balls apart), this is a combinations problem. Thus of the 15 balls, we choose. This leads to: ( ) 15 = 5005 So, there are 5,005 total samples b) How many samples contain only white balls? 1
In this question, we focus on only the white balls. So from the 8 white balls, we choose of them. This leads to: ( ) 8 = 8 So there are 8 samples that contain white balls only. c) How many samples contain two red balls and white balls? Again, order doesn t matter. So we are interested in the number of ways we can pick red balls and white balls. There are ( ( ) = 15 ways of choosing exactly red balls and 8 ) = 70 ways of choosing white balls. Now, using the multiplication principle, this leads to: ( ) ( ) 8 = 15 70 = 1050 So there are 1050 samples that contain red balls and white balls. d) How many samples contain at least red balls? This is what mainly separates section 5.5 to this one. We are dealing with cases in this question. The trigger word at least... was used, so that should be a reminder that we have cases. As you might have heard me repeat many times in tutorials, consider the compliment! It can save you time. So let s look at what we re working with in this question, without the complimentary event being applied 1. exactly red balls. exactly 3 red balls 3. exactly red balls. exactly 5 red balls 5. exactly red balls So, we would have to consider 5 cases to look at the number of ways of getting at least red balls. I.e., Look at getting red balls, 3 red balls,..., red balls. Looking at the compliment, we would take the total number of combinations, and subtract out those ways of getting: 1. exactly 0 red balls. exactly 1 red ball
There s only calculations needed to be done to find the complimentary event! So, with that being said, let s proceed with using the complimentary way. Case 1: To get exactly 0 red balls, this implies we choose white balls, of the 8. Thus, the number of ways of getting this case is ( 8 ) = 8 Case : To get exactly 1 red ball, this implies we choose 5 white balls, of the 8 and 1 red ball of the. Thus, the number of ways of getting this case is ( ( 8 5) 1) = 33 Since this is the complimentary event, we need to take the total number of ways we can select balls, and subtract out the two cases, since they are the number of ways of not getting at least red balls. So, the final answer is: ( ) 15 ( ) 8 ( ) 8 5 ( ) = 5005 8 33 1 = 1 If you got time on your hands (which I doubt many of you do), you can try the more raw method and try computing the direct way with 5 cases. Question (#, pg 5) Investment Portfolio: In how many ways can an investor put together a portfolio of five stocks and six bonds selected from her favourite nine stocks and eight bonds? In this question, order does not matter. This is simply going to take the number of ways of picking out 5 stocks from her favourite 9 and picking out bonds from her favourite 8. From here, using the multiplication principle, we get the total combinations consisting of what she desires. Thus the answer is simply: Question 3 (#9, pg 5) ( ) 9 5 ( ) 8 = 358 Poker Hands: Recall, a poker hand consists of 5 cards selected from a standard deck of 5 cards. How many poker hands consist of three cards of one denomination and two cards of another denomination? (Typically called a full house) Before we get to picking out cards, we need to understand what this question is saying. We got 5 cards which are divided into suits, 13 denominations and colours. Here, we wish to select cards by denominations. Let s do this systematically by experiment. 1. We will select one of the first denominations. Hence of the 13 denominations (Ace to King), we select 1. Hence there are ( ) 13 1 ways of selecting the first denomination. Once selected, there are replicas of it in the deck each following a different suit. Thus, we are choosing 3 of the same card of one type of denomination. Thus, there are ( )( 13 1 3) = 5 ways of choosing the first 3 cards of one denomination.. Now we need to select the second denomination. Since we have chosen one for the first cards, we need to select a different denomination. So there are 1 denominations remaining, and we 3
need to choose 1. Hence, there are ( ) 1 1 denominations to choose from for the last 3 cards. Again, once we fix a denomination, we need to pick out 3 cards of that type. Thus, we choose of the same card of one type of denomination for the last cards. This produces a total of ( )( 1 1 ) = 7 ways of choosing the last cards under the rule. Using the multiplication principle, we obtain our final answer which consists of the total number of ways we can get a full house. So there are ( ) ( ) ( ) ( ) 13 1 = 5 7 1 1 5.7 = 37 This section deals with the Binomial Theorem. Recall, the Binomial Theorem is just a quick way of carrying out some binomial expansion. Recall, it states that: (x + y) n = n i=0 ( ) n x n i y i i If you want a slightly more in-depth explanation as to how the Binomial Theorem is used, check out the third set of notes (click me). Question (#19, pg 30) How many terms are there in the binomial expansion of (x + y) 19 To do this question a bit more strategically, let s look at the binomial theorem for this question, which implies that n = 19. This leads to: 19 ( ) 19 (x + y) 19 = x 19 i y i i i=0 From here, you need to observe that i can take on values from 0 to 19. In set notation, i {0, 1,,..., 19}. Hence, n(i) = 0 (the number of elements that i can take on is 0). This implies that when you expand out the binomial expression (x + y) 19, you get 0 terms. Thus the answer is 0. In general, the binomial expansion of (x + y) n has n + 1 terms. Question 5 (#1, pg 30) Determine the first three terms in the binomial expression of (x + y) 10 The first three terms of the binomial expansion are found by subbing in the first three values i can take ( on in the Binomial Theorem. Thus, we need to let i = 1, and 3. Thus, the first three terms are: 10 ) 0 x 10 0 y 0 = x 10, ( ) 10 x 10 1 y 1 = 10x 9 y and ( ) 10 x 10 y = 5x 8 y 1 Note, don t be fooled by counting from i = 1 to i = 3. Although these are consecutive numbers, they are not the first three values you can find through the binomial theorem!
Question (#5, pg 30) Determine the middle term in the binomial expression of (x + y) 0 Recall question, to find the number of terms there are in the expansion of (x + y) 0, we simply take n and add 1. Thus there are 1 terms here. So, to find the middle term, we need to find the 11 th term of i. This is when i = 11, because there would be 10 numbers to the left and 10 numbers to the right. Note, this is the case because i goes from 0 to 0, which has 1 values. Simply letting i = 10, we can find the middle term. Thus, the middle term is: ( ) 0 x 0 10 y 10 = 1875x 10 y 10 10 In general, the middle term is found by setting i = n, given that n is an even number..1 This section was mainly used to emphasize mutually exclusive events. Recall, events are mutually exclusive when their intersection contains the null set! I.e., this means that the two events cannot happen together. This brings me to two main ideas: 1. If two events are mutually exclusive:. If two events are not mutually exclusive: n(a B) = n(a) + n(b) (1) P (A B) = P (A) + P (B) () n(a B) = n(a) + n(b) n(a B) (3) P (A B) = P (A) + P (B) P (A B) () I stated the results above because when you do probability problems revolving around Venn-Diagrams and you are given that the events are mutually exclusive, you may have adequate information to solve the problem. Just remember to underline the words independent or mutually exclusive! Question 7 (#15, pg 9) Let S = {1,, 3, } be a sample space, E = {1} and F = {, 3}. Are the events E F and E F mutually exclusive? Note, this question is asking if the events E F is mutually exclusive from E F To answer the question, let s solve it one at a time. First, let s find E F. Thus, E F = {1,, 3}. Using DeMorgan s Law, E F = ((E F ) ) = {1,, 3, }. So the events E F and E F are not mutually exclusive because (E F ) (E F ) = {1,, 3} 0. Question 8 (#19, pg 9) Genetic Traits: An experiment consists of observing the eye color and gender of the students at a certain school. Let E be the events blue eyes, F be the event male and G be the event brown eyes and female. 5
a) Are E and F mutually exclusive? E consists of the event that a student has blue eyes. So, for a student to have both blue and be a male possible So if an intersection is possible, the two events cannot be mutually exclusive! a) Are E and G mutually exclusive? Again, E consists of the event that a student has blue eyes. So, for a student to have blue eyes and have brown eyes, it is impossible (assuming students can t display central heterochromia, eyes of different colours). Thus the intersection produces an empty set. Therefore, the two events are mutually exclusive! a) Are F and G mutually exclusive? F consists of the event the student is a male. Student s can t be male and female (assuming no transgender individuals). So if the intersection is again the empty set, the two events are mutually exclusive!. The main ideas to grasp from this section was to introduce the basic ideas of probabilities and odds against/in favour. Question 9 (#39, pg 0) Convert the odds of 10 to 1 to a probability. This event is saying that for every 10 times the event could occur, the exact opposite could occur once, assuming the experiment is repeated (10 + 1) times. Thus, the probability is simply 10 = 10. 10+1 11 Question 10 (#1, pg 0) Convert the probability of 0. to odds. Interpreting this question, we see that there is a 0% chance of some event occurring. This means that if we were to repeat the event 100 times, 0 of those times the event would occur. This implies that 100 0 = 80 times the event won t occur. So as an odds ratio, this means that for every 0 times the event occurs, 80 times it won t. Thus the odds in favour are 0:80=1:, or 1 to. The odds against would be the same ratio, but reversed (:1)..3 This section covers the same word problems from 5., except we are asked to find probabilities. Note, as I have said to many people and in the tutorial, the same rules you applied to finding the number of
elements for some Venn-Diagram related problem can still be applied to probabilities. Thus, we have the inclusion-exclusion principle for probabilities which states: DeMorgan s Law for probability: Finally, we have the Compliment Rule: Question 11 (#5, pg ) P (A B) = P (A) + P (B) P (A B) (5) P (A B ) = P ((A B) ) () P (A) = 1 P (A ) (7) Balls in an Urn: An urn contains six green balls and seven white balls. A sample of four balls is selected at random from the urn. Find the probability that a) the four balls have the same colour. There are only two cases where this event could occur. Of the green balls, we choose all, or of the 7 white balls, we choose all. Thus there are cases here, where there are ( ) = 15 ways of choosing all green balls and ( 7 ) = 35 ways of choosing all white balls. Since we want probabilities, we need n(u). In this problem, there are 13 balls, of which we choose. Thus there are ( ) 13 = 715 ways of choosing 3 balls. Thus, our final answer is: ( ( 7 ) ) b) ( 13 ) + the sample contains more green balls than white balls. This could happen one of two ways: 1. All green balls, no white balls ( G, 0 W). 3 Green balls, 1 white ball ( 13 ) = 15 715 + 35 715 = 0.0993 Now, we just need to compute the probabilities of these events independently and add them. Thus, there are ( ( ) = 15 of choosing green balls. Next, there are ( 3) 7 1) = 10 ways of choosing exactly 3 green balls and 1 white ball. Dividing both cases by n(u), we obtain the following answer: ( ( ) ( ( ) 3 7 13 ) + 1) 15 + 10 ) = 715 ( 13 = 0.178 7
Question 1 Dice: A dice is rolled 7 times. What is the probability of rolling something greater than a at most 3 times? Let R = the event that something greater than a is rolled. R = D = the event that something less than or equal to a is rolled We are interested in looking at the cases where event R happens at most 3 times. This means that it could happen the following ways R doesn t happen at all R happens exactly once in 7 rolls R happens exactly twice in 7 rolls R happens exactly 3 times in 7 rolls Next, we need to assign probabilities for event R and D. The number of ways we can get something greater than a on a die is getting a 5 or. This is possible ways of the, thus P (R) = = 1. Using 3 the Complimentary Rule, P (R ) = P (D) = 1 P (R) = 1 1 =. Going back case by case, we get the 3 3 following probabilities: 1. R doesn t happen at all. This means that all rolls correspond to outcome D. This has a probability of ( 3 )7 = of occuring.. R happens exactly once in 7 rolls. This means that event R happens once, and D happens 7 1 = times. Among these 7 rolls, R could occur in any 1 of 7 rolls. Thus, we need to take account of all these cases and multiply the probability by ( 7 ). So, our answer for this case is: ( ) 7 1 ( ) 1 3 3 3. R happens exactly twice in 7 rolls. Much like the last case, this means that even R happens exactly twice, and D happens 7 = 5 times. Among these 7 rolls, R could occur twice. Thus, we need to again consider all cases and multiply this probability by ( 7 ). So the answer for this case is: ( ) 7 ( ) 1 3 ( ) 5 3. The final case is simple to understand once you get the previous two and is ( ) ( ) 3 ( ) 7 1 3 3 3 Summing up the probabilities associated with each possible case above, we get the following result: 0.8703. 8
Question 13 Poker Hands: What is the probability that you obtain a poker hand consisting of only Kings and Queens?. Note, a poker hand is a 5 card hand. Before we do anything fancy, let s just address the total number of samples possible for a poker hand. Of the 5 cards, we simply choose 5 (order doesn t matter). Thus there are 5 = 59890 samples. 5 Next, let s look at all the possible cases and their corresponding probabilities. 1. Kings, 1 Queen. To get Kings, we need to pick Kings of the Kings. To get 1 Queen, we need to choose 1 Queen from Queens. Thus there are ( ( ) 1) = ways of this event occurring with probability. 59890. 3 Kings, Queens. To get 3 Kings, we need to pick 3 Kings of the Kings. To get Queens, we need to choose Queens from Queens. Thus there are ( ( 3) ) = ways of this event occurring with probability. 59890 3. Kings, 3 Queens. To get Kings, we need to pick Kings of the Kings. To get 3 Queens, we need to choose 3 Queens from Queens. Thus there are ( ( ) 3) = ways of this event occurring with probability. 59890. 1 King, Queens. To get 1 King, we need to choose 1 King of the Kings. To get Queens, we need to choose Queens from Queens. Thus there are ( ( 1) ) = ways of this event occurring with probability. 59890 The answer we desire is just the sum of all the cases mentioned above resulting in a probability of +++ = 5 = 0.000155 59890 59890. This section covers independent events and conditional probability. Recall, two events are independent if Conditional probability implies that: P (A B) = P (A) P (B) (8) P (A B) = P (A B) P (B) (9) From experience, when you see words like given that... or if it is known that..., you should be thinking about conditional probabilities! 9
Question 1 a) This is more of a thinking question. Can two events be both mutually exclusive and independent, if P (A) > 0 and P (B) > 0? Recall, if two events are independent, then P (A B) = P (A) P (B). However, if two events are mutually exclusive, then P (A B) = 0. But, it was stated in the question that P (A) > 0 and P (B) > 0. So, P (A) P (B) > 0, therefore they cannot be both mutually exclusive and independent in this case. b) When can two events be both mutually exclusive and independent? Again, if two events are independent, then P (A B) = P (A) P (B). However, if two events are mutually exclusive, then P (A B) = 0. Thus, if P (A) = 0 or P (B) = 0, then it is possible to satisfy the condition of independence and be mutually exclusive at the same time! In short, two events are both mutually exclusive and independent if and only if one of the events has a probability of 0! Question 15 (#1, pg 7) More Balls in an Urn: Two balls are selected at random from an urn containing two white balls and three red balls. What is the conditional probability that both balls are white, given that at least one of them is white? This question is very similar to your assignment! Looking at some trigger words, we see the words given that.... We should be thinking conditional probabilities. So, let A be the event that at least one ball is white. B be the event that both balls are white Thus, we are looking for P (B A) = P (B A) P (A) The tricky part in this question is to understand that B A. this implies that P (B A) = P (B) So, to find P (B), we need to simply find n(b) and divide it by n(u). However, it is clear to see that n(u) = ( 5 ) = 10 so, n(b) is ( ) n(b) = P (B) = n(b) n(u) = 1 10 = 0.1 10
Next, we need to find P (A). being a bit more smart, as we always are, we can find the complimentary event. If A is the event that at least 1 ball is white, then A is the event that no ball is white. This means both balls are red. Thus, ( ) 3 n(a ) = = 3 P (A ) = n(a ) n(u) So, using the complimentary rule, we get that = 3 10 = 0.3 P (A) = 1 P (A ) = 1 0.3 = 0.7 Finally, subbing this all back into our original equation we get Question 1 P (B A) P (B A) = P (A) = P (B) P (A) = 0.1 0.7 = 1 7 Suppose we have independent events A and B with P (A) = 0.3 and P (B) = 0.. Find P (A B ). The most important part of this question is to understand that the events are independent. This means that P (A B) = P (A) P (B). Thus, P (A B) = P (A) P (B) = 0.3 0. = 0.18 Next to find P (A B ), let s use DeMorgan s Law. So P (A B ) = P ((A B) ) = 1 P (A B) = 1 0.18 = 0.7 11
Question 17 (#, pg 77) Roll a Die: Roll a die and consider the following two events. E = {,, } and F = {3,, }. Are the events E and F independent? Just writing things in point form, we get P (E) = 3 P (F ) = 3 P (E F ) = n(e F ) n(u) = 1 3 Since P (E F ) P (E) P (F ), these two events are not independent! 1